Answer: In KMnO4, Mn is in +7 oxidation state and have all the 3d-orbitals vacant. Mn7+ ion is surrounded tetrahedrally by four oxide ions. All oxide ions have filled 2p-orbitals. There is transfer of an electron from filled 2p-orbitals of oxide ion to vacant d-orbitals of Mn7+ ion. Since p-orbitals are ungerade and d-orbitals are gerade, therefore, electron transition from p-orbitals of O2- to d-orbitals of Mn7+ ion is Laporte allowed i.e., ΔL = ±1 and also there is no change in spin multiplicity during electronic transition. Therefore, transfer of an electron is Laporte and Spin allowed. Therefore, KMnO4 is intensely purple in colour.
Why is [Ti(H2O)6)] 3+ violet in colour. The violet/purple colour of [Ti(H2O)6)] 3+ ion is due to absorption of blue-green light and transition of electron of the t2g orbitals to one of the degenerate eg ortibals. The absorption spectrum of [Ti(H2O)6)] 3+ reveals that the d-d transition occurs with a single broad peak.
What is the reason for the colour of potassium dichromate. Colour of dichromate ion. Why is chromate yellow in colour. Charge transfer transition. LMCT transition.
What is the difference between d-d transition and charge transfer transition. How do we distinguish between d-d and MLCT transistions.